MHT CET · Physics · Dual Nature of Matter
A metal surface of work function \(1.13 \mathrm{eV}\) is irradiated with light of wavelength \(310 \mathrm{~nm}\). The retarding potential required to stop the escape of photoelectrons is [Take \(\frac{\mathrm{hc}}{}=1240 \times 10^{-9} \mathrm{SI}\) units]
- A \(1 \cdot 13 \mathrm{~V}\)
- B \(2.87 \mathrm{~V}\)
- C \(3.97 \mathrm{~V}\)
- D \(4.23 \mathrm{~V}\)
Answer & Solution
Correct Answer
(B) \(2.87 \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
Energy of incident light:
\(\begin{aligned}
E & =\frac{h c}{\mathrm{e} \lambda} \\
& =\frac{1240 \times 10^{-9}}{310 \times 10^{-9}} \\
& =4 \mathrm{ev}
\end{aligned}\)
Stopping potential \(\mathrm{V}_0=\frac{\mathrm{hc}}{\mathrm{e} \lambda}-\frac{\phi_0}{\mathrm{e}}\)
\(\mathrm{V}_0=4-1.13=2.87 \mathrm{~V}\)
\(\begin{aligned}
E & =\frac{h c}{\mathrm{e} \lambda} \\
& =\frac{1240 \times 10^{-9}}{310 \times 10^{-9}} \\
& =4 \mathrm{ev}
\end{aligned}\)
Stopping potential \(\mathrm{V}_0=\frac{\mathrm{hc}}{\mathrm{e} \lambda}-\frac{\phi_0}{\mathrm{e}}\)
\(\mathrm{V}_0=4-1.13=2.87 \mathrm{~V}\)
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